INFINITY SOLUTIONS

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MATHPRITOM
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INFINITY SOLUTIONS

Unread post by MATHPRITOM » Sun Feb 05, 2012 7:45 am

Prove or disprove that, the equation $ x^2+y^2+z^2 =59^n $ has solution for every positive value of n.

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Phlembac Adib Hasan
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Re: INFINITY SOLUTIONS

Unread post by Phlembac Adib Hasan » Sun Feb 05, 2012 8:21 am

Yes, proved.
Well,it's easy for odd $n$.As \[7^2+3^2+1^2=59^1\]So after multiplying both sides by $59^2$ we find it is true for $n=3$ and so on.The even case is also true as \[58^2+9^2+6^2=59^2\]
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sm.joty
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Re: INFINITY SOLUTIONS

Unread post by sm.joty » Sun Feb 05, 2012 3:27 pm

Well here is my proof.
It's easy to see that $x,y,z$ must be odd.Because rest of the cases are disable for this equation.
Now $x^2=8k_1+1$
$y^2=8k_2+1$
$z^2=8k_3+1$
$k_1,k_2,k_3$ positive integers.
Then from the equation we have,
$k_1+k_2+k_3=\frac{59^n-3}{8}$
It's easily provable for all odd $n$
$59\equiv3 (mod 8)$
so,
$59^n\equiv 3^n\equiv 3 (mod 8)$

for all even $n$ the base case is true.(like Adib's example)
By induction hypothesis we assume that
$x^2+y^2+z^2=59^{2n}$ have infinite solutions.
multiplying both sides by $59^2$ we get,
$(x^2+y^2+z^2)59^2=59^{2(n+1)}$
$(59x)^2+(59y)^2+(59z)^2=59^{2(n+1)}$

Thus it's true for all even $n$

So this equation have infinitely many solution. :mrgreen:
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MATHPRITOM
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Re: INFINITY SOLUTIONS

Unread post by MATHPRITOM » Sun Feb 05, 2012 10:04 pm

sm.joty wrote:Well here is my proof.It's easy to see that $x,y,z$ must be odd.
Let,n=2 ,so, the equation is $x^2+y^2+z^2=59^2$.
Now, $59^2=(2.5^2+3^2)^2=2^2.5^4+3^4+2.2.5^2.3^2 $
Now,let,$ x^2=2^2.5^4,y^2=3^4,z^2=2.2.5^2.3^2 $ so,I don't understand why x,y,z must be odd?

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afif mansib ch
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Re: INFINITY SOLUTIONS

Unread post by afif mansib ch » Mon Feb 06, 2012 2:00 pm

@mathpritom via,
\[59\equiv 3(mod4)\]
for n even and odd respectfully \[n^2\equiv o(mod4)\]
or\[n^2\equiv 1(mod4)\]
but\[x^2+y^2+z^2\equiv 3(mod4)\]
can only be possible when \[x^2,y^2,z^2\equiv 1(mod4)\]
so x,y,z are odd.

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Phlembac Adib Hasan
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Re: INFINITY SOLUTIONS

Unread post by Phlembac Adib Hasan » Mon Feb 06, 2012 3:16 pm

@ Mansib vaia, this proof works for odd $n$s.But for even $n$, all of $x,y,z$ can't be odd as $59^{2k} \equiv 1(mod\; 4)$.Actually two of them must be even and the other will be odd.(Like $58,9,6$)That's why Pritom Vaia was confused.I think Joty Vaia should replace the sentence "It's easy to see that x;y;z must be odd" by "It's easy to see that for odd $n$s,x;y;z must be odd".
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sm.joty
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Re: INFINITY SOLUTIONS

Unread post by sm.joty » Mon Feb 06, 2012 11:08 pm

Phlembac Adib Hasan wrote:@ Mansib vaia, this proof works for odd $n$s.But for even $n$, all of $x,y,z$ can't be odd as $59^{2k} \equiv 1(mod\; 4)$.Actually two of them must be even and the other will be odd.(Like $58,9,6$)That's why Pritom Vaia was confused.I think Joty Vaia should replace the sentence "It's easy to see that x;y;z must be odd" by "It's easy to see that for odd $n$s,x;y;z must be odd".
Humm...... got the mistake.
Actually I should replace that "two of the $x,y,z$ is even or all of them is odd."
But the think is that we can just use induction for both odd and even $n$
There is no need to use Modular. :D
So I should use induction.
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Masum
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Re: INFINITY SOLUTIONS

Unread post by Masum » Fri Feb 10, 2012 6:12 pm

MATHPRITOM wrote:Prove or disprove that, the equation $ x^2+y^2+z^2 =59^n $ has solution for every positive value of n.
Note that $59=5^2+5^2+3^2$. Then it immediately follows from the known fact that The product of two numbers of the form $a^2+b^2+c^2$ is of the same form.(Prove yourself)
By the lemma, we can conclude that any power of $5^2+5^2+3^2$ is of the same form $x^2+y^2+z^2$. Thus proved.
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